Differential Equations Solution of the Heat Equation

In this lesson, our instructor Will Murray gives an introduction on solution of the heat equation. He discusses the partial differential equation and how to solve it, as well as the procedure for the heat equation.

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I have watched both the probability and differential equations courses and they are both great! You explained the concepts very well. Question. Do you have any plans to create a new course for Educator.com in the future? If so, I would like to make a suggestion. Math for quantum mechanics. thanks

regards,

Joe

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Last reply by: Dr. William MurrayMon Aug 17, 2015 10:10 AM

Post by Muhammad Asad Ullah MOAVIAon August 13, 2015

Dear Professor William Murry, Thank you so much for providing an amazing mathematics courses. I was very weak in differential equation and in probability and I have solved all my problems through your courses. I will always remember you in my prayers! Dear Professor, I have one question to you. If you provide a real analysis course through educator.com, it would be very nice of you. As I know you are connoisseur in the field of mathematics, at least you would guide me in real analysis course. As for me, I am doing my M.Phil in economic policy. I am keen to learn the game theory, so for game theory is concerned, it is all about real analysis and optimization. I am good in optimization, but very poor in real analysis. I am looking forward to hear from you soon.Cordially, Assad Turkey

1 answer

Last reply by: Dr. William MurrayWed May 6, 2015 8:54 PM

Post by Alexander Guevaraon May 4, 2015

HI Dr. Murray, do you have tutorial on Sturm Liouville?

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Last reply by: Dr. William MurraySun Jan 4, 2015 7:41 PM

Post by John Nashon December 31, 2014

Happy New Year Sir,

You have been of great help for my differential equations course I have taken this semester.

1 answer

Last reply by: Dr. William MurrayMon Dec 15, 2014 1:19 PM

Post by suzanne El Shafeion December 12, 2014

Hi there,

I have a question I am trying to solve but my initial conditions are f(x)=100x 0<x<2 =100(4-x) 2<x<4

How would you suggest i set this up for the integration can i have an integral over 100x only?

1 answer

Last reply by: Dr. William MurrayTue Dec 10, 2013 11:47 PM

Post by Edward Arreguinon December 9, 2013

Can you make videos on the solution for the insulated rod and the steady state solution?

1 answer

Last reply by: Dr. William MurrayWed Nov 20, 2013 9:55 AM

Post by Satendra Guruon November 12, 2013

I am confussed when you arrive at the 2n+1 for odd numbers? Can you explain this better.

Thanks,

Solution of the Heat Equation

Solution of the Heat Equation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Transcription: Solution of the Heat Equation

Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and we are studying partial differential equations.0000

What we are going to do in this lecture is solve the heat equation we are going to combine everything we learned about Fourier series and separation of variables, put it all together and get an actual solution to the heat equation.0009

Let us see how that works out, we are going to solve the heat equation and its boundary and initial conditions, I will just remind you that this was the PDE that we started with.0021

PDE stands for partial differential equation and we had these two boundary conditions U(0t)=0 and U(Lt)=0 and those are the boundary conditions and then we had this initial condition which told us what happened when we plugged in t=0.0030

That represented the initial distribution of heat along this rod, let us see what the progress that we have made so far, we started to solve this using separation of variables that was in an earlier lectures, you will see one called separation of variables.0053

We did a lot of work back in this earlier lecture and we got a general solution to both the partial differential equation and it also match the boundary conditions.0070

And this what we got U(Xt)= a series b(n) x sin(n) π(x)/L and then this fairly complicated expression e^-(n^2 π^2 α^2)t/L^2.0082

This looks very complicated but it is actually very formulaic, in particular all of these constants comes straight out of the differential equation, for example the α^2 comes straight from the differential equations, you can just drop that in.0098

Same with the L, straight from the differential equations, you can just drop that in, this is very formulaic, all you have to do is just drop in the constants, the only part here that we have not figured out yet is the (bn).0113

That is the part they we are going to focus on in this lecture, what we try to do here is we try to match the initial condition, I said we already got a solution, this solution the we just found is it already matches the partial differential equation.0128

And it matches the boundary conditions, the only part that it does not necessarily match yet is the initial condition, which is that U(x0) should be equal to f(x), let us see what that means when you plug in t=0 into our solution.0143

That means you are plugging in 0 here, which means you are just getting e^0 which is 1, all of those e terms just turn into 1 and what we got is the sum of (bn) x sin(n π(x)/L)=f(x).0160

What you see here is that is exactly a Fourier series, that is why we did all that work in the previous lecture learning about Fourier series is because we are going to use it right now, we need to expand f(x) into a Fourier series.0177

We want a Fourier series which uses only sines and we want to find those coefficients (bn) by finding a Fourier series for f(x) that uses only sines, let me show you what the procedure is.0195

Remember that you are given f(x), it will be defined between 0 and L because that represented the initial distribution of heat in this rod and we assumed that the left hand under the rod was 0.0211

The right hand end of the rod was L and that f(x) is defined between 0 and L, what we want to do is extend it to -L, and extended to be an odd function and the reason for that is that then we know that it is a Fourier series we will have only sines.0230

That is something we learned out in the previous lectures, if you have not just watch the previous lecture on Fourier series, this is where we are using it, you might want go back and check that previous lecture.0251

Then we are going to find the Fourier series for f(x) and we figured out when you have an odd function, it will be easier to find the Fourier series that it is for general functions, just have this (bn) is equal to 2/L x the integral from 0 to L of f(x) x sin((n π(x)/L(dx)).0260

That tells you what the (bn) is, all you do is you take that (bn) and you drop it back in to our general solution to the heat equation, you just take this BN, drop it back in right here and everything else like I said is predetermined from the differential equation.0281

You got the L, the α^2 from the differential equation, everything else just comes straight out the differential equation, the only work here is finding the (bn).0298

You are going to have to do some integration to find the (bn) but everything else just follows from all the work we have done in the previous lectures, let us go ahead and see how those work out.0307

I broke it up into several steps over various examples here, we are not going to solve the whole thing from scratch in the first example, we are just going to start with the boundary value problem.0322

What are we doing in this first example is we will extend the initial function f(x) to be an odd function.0333

The initial function is f(x)=3 that is define from 0 to L=3, let me make my graph and there is 3, there is 0, there is -3 and the function we are given is just as f(x) is equal to 3, between 0 and 3.0342

What we want to do is extend that to be an odd function, remember odd functions have rotational symmetry, if you rotate them 180° around the origin then they should look the same.0369

What we are going to do is extend that from -3 to 0 and we will extend it in such a way that it has rotational symmetry, we will define f(x) to be equal to, well we are given that it is 3, if 0 is less than or equal to x , is less than or equal to 3.0385

We will extend it to be, there is 3, there is -3, we will extend it to be -3, if -3 is less than x, less than 0, that way f(-x)=-f(x), we will get an odd function there.0409

The point of doing that of course is that later on it is Fourier series we use only sines, this is an odd function.0437

Just a recap here, we are not completely solving our partial differential equation yet, we are just doing the first step which is to take the initial function and extend it, that is the initial function we are given and there is our extension.0448

Extending it in such a way that it will be an odd function, an odd function has rotational symmetry that means if it is +3 on a positive set of x values, we are going to make it -3 on the negative set of x values.0464

This 3 was given to us, this -3 is our contribution in this problem, example 2, we are going to keep going with a function we justified in example 1, this all came from example 1.0481

What we want to do here is to find a Fourier series that uses only sines, let me remind you of our formula for Fourier coefficients, BN is equal to 2/L, integral from 0 to L and of f(x) x sin(n π(x)/L)d/x.0505

Let us work that out for this function, our (BN) is equal to, our L =3, that is our L right there, our (BN) =2/3 x the integral from 0 to 3.0532

Our f(x) between 0 and 3 is just 3 x sin(n π(x/L), our L is still 3, (dx), we just have to work out this integral which is not too bad, it is 2/3, actually the 2/3 and then we still have another 3 from the inside.0551

The integral of sine is -cosine, -cos of, I should have put n(π x/3), n(pix/3) x 3/n(π), that is the chain rule coming through there and then we have to evaluate this from x=0 to x=3.0580

Let me simplify my coefficients a bit, 2/3 x 3 is 2 and then that times 3 is a 6/n(π), if we plug in x=3 there, we have -cos(n π) - -1, +1.0612

1 - cos(n π) there and we figured out in the previous lecture a little formula for cos(n π), I figured it out real quick right now, cos(0) is 1, cos(π) is -1, cos(2pi) is 1, cos(3pi) is -1.0650

cos(n π) is 1 if n is even and -1 if n is odd, what we have here is 6/n(π), now -cos(n π) if n is even then cos(n π) will be 1, we will get 1 - 1, this will be 0 if n is even.0677

If n is odd then this would be -1, 1 - -1, 1 + 1 would be 2, if n is odd, of course we are multiplying that by 6/n π, this is all equal to, if I have to write a couple cases here 6x 2 is 12/n(π).0718

If n is odd, still 0 if n is even, our Fourier series of x, let me remind you of the general formula for Fourier series, I'm not going to write the (an) and the cosine terms because we know we are dealing with an odd function here.0745

There will be no cosine terms, our Fourier series of x, I'm just going to write the sign terms n=1 to infinity(bn) sin(n π) x/L, that is our generic form for a Fourier series.0770

Let me plug in the (BN ) that I just arrived here, b1 is 12/π, 12/π x sin of n is 1 π(x)/3, L is 3, b2 is 0, I will go ahead and write the 0, B3 is 12/3 π, I do not think I will simplify that, I think it will be easier to see a pattern.0790

sin(3pi(x)/3) + 0 + 12/5pi sin(5pi)x/3 and so on, we just write that 1 more time without the 0s, 12/π sin, I think I could factor out a π, maybe I will leave it in there.0824

sin(π x/3)+ 12 over 3 π, sin(3pix/3) + 12/5pi, sin(5pi/3) and that is certainly enough to get the pattern and see how it would continue from there, that is our Fourier series for the function.0855

Let me recap how we derived that, we use our generic Fourier series formula BN is equal to 2/L times the integral from 0 to L of f(x) sin(n pix/L), that is our Fourier series formula when we know that the function is odd.0892

Of course we define this function to be odd because we wanted to get a sine series, that is our (BN) and of course we know that the (an's) are all 0 for odd functions, we don't have to worry about the An's.0911

I just plugged in my values here, L is 3, there is the L everywhere there, now this three came from this three right here, that was our definition of F, we had to do an integral there, we get 2/3 × 3 which was 6, 2/3 × 3 was 2.0924

And then an extra three and there is what gave us the 6, the integral sin(n π(x)/3) is negative cosine of n(π)x/3, but we got a multiply by 3/n(π), that's why that 3/n(π) came out of.0948

I plugged in X equals 3, we got -cos(n π )plugged in X equal 0, we get the cosine of 0 is just 1, we get 6/n(π) x over and high times the negative cosine n π +1 and I was trying to remember a formula for cosine(n π).0967

When we came up with this it is -1 if n is odd, 1 if n is even and if N is even, the 1s cancel each other out and we just get 0, if n is odd you get 1 minus -1, we get 2 here and BN is 2 times 6/n π, 12/n(π) for odd n's and 0 if n is even.0983

Now that's my (BN), I'm going to drop that back into a Fourier series which means I take those coefficients and I just multiply each 1 by sin(n π)x/L, the even ones all give me 0 and the odd ones give me 12/n(π).1010

There is 12/1(π), 12/3(π), 12/5(π) and so on, we get just by simplifying that 12/π times sin(pix)/3, 12/3pi sin(3pi)x/3, 12/5(π) x sin(5pi)x/3, that is my Fourier sin series for that initial function.1026

We still have not solve the partial differential equations but we are very close now that we got the (BN), essentially we are just going to drop them into our generic formula and we will have the solution to the partial differential equation.1048

In example 3, we are actually going to solve the boundary value problem starting with the partial differential equation, UT=α^2 Ux(x), that is our heat equation.1063

We had 2 boundary conditions and we have 1 initial condition and we done all the work for this already, we really do not need to do any more work in this 1 we just need to take our answers and drop them in to our generic formula.1074

I will remind you what we figured out in example 2, I believe it was, I you have not just watched example 2, maybe check back and look at example 2 and you'll see that we figured out that our (bn) we are (bn) was 12/n π.1094

If n was odd, just 0 if n is even and we can just take those BN's and plug them into our general formula for the solution of the heat equation, it is a little messy but it is not hard at all.1113

Our general formula is U(XT) is equal to the sum from n= 1 to infinity of (BN) sin(n π)x/L x e^-(n^2 π^2 α^2t)/L^2, what have we been given here we been given that L is equal to 3.1133

Where will I see an L, I'm going to plug in 3 and we have not been given α ^2, I'm just going to leave that as α ^2, but we have been given what BN is, well we have not been given that if we figured it out in example 1.1169

Let me drop in my (BN) from example 2 that is, let me drop in my (bn) from example 2, all the even ones are 0 and I'm going to fill in the odd ones here, this is for n=1, we get 12 /π sin(π)x/3, e^-π ^2 α ^2 T / 9 because that L^2.1187

Now n= 2 gives us 0 for n equal 3 we get 12/n π, 12/3pi sin(3pi)x / 3 x e^-, n^2 is 9, I'm just going to leave that as 9 even though it does cancel, but I think it is easier to spot a pattern if we do not cancel it.1218

9pi ^2 n^2 π ^2 α ^2 t / L^2 was 9 and the next even term is still 0, 12/5pi sin(5pi)x/ 3 x e^-20 π^2 α^2t / 9 and it keeps going like that.1247

I see that I got this pattern of odd numbers building up 1, 3, 5 and in my e's I got 1, 9 and 25, those are the squares of the odd numbers, I know there is a way to keep track of odd numbers which is to keep track of 2n + 1.1278

If you start summing from n=0 , 2n+ 1 will give you all the odd numbers, let me we write a close form expression for that.1297

The sum from n equal 0 to infinity of 12/the odd number x π, 2n + 1 x π sine of odd number x π/x 2n + 1 x π/x/ 3, 3 in all of those, e^negative, now odd number ^2, 2n + 1^2 π^2 α ^2t and there was a 9 in all of those and that is what I have.1304

Congratulations if you made it this far with me because together we have solved our very first heat equation, our very first partial differential equation took for it 4 lectures to do it, it is quite a lot of work we went into that.1354

But at the very end it is pretty straightforward because we done all the work in previous examples, in particular in example 2 we found our (BN) was 12/n π if n is odd, 0 if n is even and we had this generic formula.1372

This is coming out of the previous lecture, it was coming out of 2 lectures ago, lecture on separation of variables, we had this generic formula for the solution of a heat equation was (bn) sin(n π)s/L e^-n^2 π ^2 α^2 T/L^2.1390

We know all those constants, we do not know the α ^2 because it was just left generic in the equation but we know that L is equal to 3 and we figured out the (bn) in example 2, I drop those in here.1422

There is my n= 1 term, n=2, in fact all the even ones are 0, they are sort of a missing 0 in each of these where I just dropped out all the even terms, there is n= 3, there is n=5.1436

And then I figured out that I can index my numbers differently if I use 2n + 1 to keep track of odd numbers then I can add them up for n=0 again, wherever you see an odd number above I change it to 2n + 1.1449

That at long last really is the solution to our very first heat equation, in the next series of examples, we are going to build up the solution to another heat equation and then we are going to draw a big sigh of relief.1495

Let us keep going here, in example 4 we are going to consider the boundary value problem below, UT(t) is equal to 4U(xx), by the way that means that the four is α ^2.1510

We got a couple boundary conditions here, U(0T) is 0, U(4t) is 0, by the way that means that L is equal 4 and U(x0) is equal to x, that is our initial condition and that is defined from 0 to 4, 0 to L.1527

The first part of solving this boundary value problem is just to extend the initial function to be an odd function, that is all we have to do right now and then in the next couple of examples we will put the other elements together.1555

The Fourier series in the general solution to get a complete solution to this partial differential equation, right now we just have to extend the initial function to be an odd function and that really not very hard.1569

Let me graph what we are looking at, this is 4, this is 0, the function we are looking at is f(x) = x from 0 to 4 and we wanted to be in odd function which means we are going to extend it on to the domain going back to -4.1585

Since we remember odd functions have rotational symmetry around the origin and we want to define this in such a way that gives us that rotational symmetry around the origin.1609

It is easy to see that if I just extended as a line back like that then that would give me that nice rotational symmetry and of course that line back there is also the line in y=x.1623

All I have to do is define f(x) to be x and instead of going just from 0 to 4, I'm going to run it now from -4 to four.1637

That is my extension and that is an odd function, all we did there, we have not solve the differential equation, we have not gone through the Fourier series business yet, we are going to do that in the next example.1652

We are going to solve this problem in the next example, all we did at this point was we looked at that initial function there and we know we want a Fourier sine series at the next step.1672

But at this step we just want to extend that to be an odd function, when we do find the Fourier series we will get all sines, we just looked at the graph of it f(x) = x.1686

And then in order to extend it to have rotational symmetry, we just extended it back and it turned out that it was still just y=x.1698

Instead of defining it from 0 to 4, I can define it from -4 to 4 and I will get my odd function ,I'm going to carry that over, we are going to use this same boundary value problem for the next couple of examples to find a Fourier series.1709

And then to use the Fourier series to actually solve the partial differential equation, the whole boundary value problem, let us look at example 5, for example 5 we are going to find a Fourier sine series for the initial function from the example above.1724

The nice thing is that we got an odd function here, we defined that back in example 4 to be an odd function, I know that I'm going to get of a Fourier series that involves only sines, it means I do not have to worry about the cosines at all know.1742

No (an) just (bn), let me remind you of the formula for (bn), BN when you have an odd function is 2/L x the integral from 0 to L of f(x) x sin(n π)x /L(DX) and we already figured out what L is.1758

Let me go ahead and drop that in our (BN), our L was 4 and 2/4 is better known as 1/2, the integral from 0 to 4, f(x) is just x, so x x sin(n π)x /4 DX, that is my (BN).1788

By the way you might remember another formula for Fourier series, where we had (BN) is equal to 1 / L the integral from -L to L and then the same stuff after that, the reason we simplify that is because we know that f(x) is an odd function.1814

If f(x) is an odd function, we have some symmetry, which can run that integral from 0 to L and multiply it by 2, now I want to solve that integral of course that means I have to do integration by parts.1831

If you are rusty on integration by parts we got some lectures in a different series here on www.educator.com, there is a calculus 2 series, it is also called the calculus BC where we have a whole lecture on integration by parts.1846

I'm assuming that you already know integration by parts, I'm going to go through a kind of quickly using the little shorthand trick that I covered in those lectures back in the calculus 2.1862

It is called tabular integration sin X x sin(n π)x/4, tabular integration you write derivatives on the left, derivative X is 1, derivative 1 is 0 and then integrals on the right and the integral of sin(n π)x/4 is -cos(n π)x/4.1873

It is -cos(n π)x/4 except that I have to multiply that by 4/n(π) by the chain rule, we multiply that by 4/n(π), my π got a little messy there.1909

4/n π and then I had to do 1 more integral there, the integral of cos is sin(n π)x/4 but I still have a negative here and I have to multiply by another 4/n(π), I get 16/n^2 π^2 and then I write this little diagonal lines.1939

Alternating signs +, - and what I see here is I got a 1/2, that still same half from the outside and I got -4x/n(π), I'm multiplying along the diagonal lines and this tabular integration.1971

4x/n(π) cos(n π)x/ 4 minus -, let us see, I forgot my negative sign here there should be a negative here, +16 / n^2 π ^2 sin(n π)x/4, I multiply along that diagonal line there and I have to evaluate this from x=0, 2X = 4.1988

That is going to be a little bit messy but it is what is nice about it is a lot of the terms go way quickly, in particular this sin term if we plug in x=4 in the sin term, we will get sin(n π) that is just 0 and if we plug inx=0 we will get sin(0) that is just 0.2043

The sin term is going to go away to 0 on both limits there, this is pretty nice, I'm going to combine my 1/2 of my -4x /n(π), I get -2/n(π).2062

Now I have got a cos( n π)x/4, that cos(n π) when I plug in x=4, that cos(4pi)x/4n(π)/4, cos(n π), when I plug in 0, there is also an X there, I also have a 4 there, when I plug in x=0, that x turns it in to 0.2083

We just get -0 there and then I said before, the sin terms will give you 0, those dropout right away and it looks like what I have here is -2 x 4 is -8/n(π).2133

Now cos(n π) we figured out a nice formula for that, cos(n π),let us see, let me write down the first couple terms there, cos(0) is 1, cos(π) is -1, cos(2pi) is 1 again, cos(3pi) is -1.2151

They are alternating between 1 and -1, cos(n π) is -1^n, -a/ n(π) x -1^n and that is my formula for BN and I guess I could simplify that a little bit by combining that negative in with the -1^n.2181

That is -1^n + 1 x 8/n(π), my Fourier series, our general formula for Fourier series, I will go ahead and write the whole formula, a0/2 + the sum from n=1 to infinity of (an)cos(n π)x/L, n(π)x/L + (bn)sin(n π)x/L.2205

However when we we said when we have an odd function and we designed this function to be odd, that makes all the the cos terms dropout which means all the a terms dropout.2248

a0 is 0 and all the cos terms are 0 because F(x) is an odd function, that means we just have to focus on the BN's and we figured out our BN it is -1^n+1 8/n(π).2261

I can factor out the a/π, I think I can do that, I will get a/π x the sum from n =1 to infinity, the rest of my (bn) I can not factor it out because it is got n's in it.2294

-1^n +1/n, and sin(n π)x, my L was 4, n(π)x over, I'm going to go ahead and change that L to a 4 and that is my Fourier series for F(x), still have not used it to solve with the differential equation yet.2312

We got a pretty close we would not have to do very much work to use this to solve the differential equation, let me just recap the steps that we did to get this Fourier series, we start out with a function F(x) = x which is an odd function.2339

Which means I know when I do my Fourier series, I would not have to do my general formula here, 1/ L x the integral from -L to L, I can use my specialized formula 2/ L x the integral from 0 to L.2354

It is such a little bit easier, makes it a little bit less messy, I plug in L=4, that is where I got 4 here, 2/4 is where I got that 1/2 here, L =4 here.2368

That x came from this x right here, that is where we got that x and then clearly that is the integration by parts problem, remembering what I learned from the integration by parts lecture in the calculus 2 lectures here on www.educator.com.2384

I set up my tabular integration x and sin(n π)x/4, wrote my derivatives here, wrote my integrals on the side, set up alternating signs positive and negative, multiply down diagonals and got this horrific expression.2404

Which turned out not to be so horrific because I know when I plug-in X =4 and x= 0, all the sin terms go to 0, the cos terms, well the X = 0 term, the cos term goes to 0 because of that x.2421

The X =4 term, that half and that 4/n(π) that is how I got the 2/n(π) and the X is where I got that 4 right there, that is where that 4 came from and we got cos(n π), of course that came from there.2438

And cos(n π) I figured out this nice formula for cos(n π) is just -1^n, this turn into the 2 and the 4 gave me 8 and I get -1^n x -1 gives me -1^n + 1 x 8/n π and that is my BN for my Fourier series.2458

Here is the generic formula for Fourier series and I do not have to worry about any of the A's because F(x) was an odd function, we extended f(x) to be an odd function precisely because we wanted to get a Fourier series involving only sines.2482

That means all of the A's disappear and go to 0 and then we just drop in what we figured out for BN's, I drop that and right there and I factored out the a/π, I can pull it out because it does not have an n in it and I just copy down sin(n π)x/L=4.2498

That is my Fourier series for the initial function F(x) still have not solved the partial differential equation the, full boundary value problem but I'm very close here, essentially there is no more work to do, I'm just going to carry these coefficients over from the Fourier series.2517

And drop them into the generic solution of the boundary value problem, that is our last example, let us go ahead and jump to that, example 6 we will solve the boundary value problem that we first met in example 4.2534

We did most of the work here in examples 4 and 5, example 6 we are just going to drop in the answers the we got from the 4th, let me remind you what we figured out earlier.2550

We figured out in example 5 it was, that we found those coefficients bn and what we got was -1 to the n +1× 8/n(π).2564

Let me remind you of the general solution to the heat equation which we worked out a couple of lectures ago but I reminded you at the beginning of this lecture what the general solution was, that is fairly complicated formula.2587

Some from n=1 to infinity of BN, it looks like the Fourier series for sin(n π)x/L but then there is also this t term, e^-n ^2 π^2 α^2t/L ^2 , rather complicated but the good news here is we already figured out all the numbers here.2606

We really do not have to do any more work, we are just plugging things in, what we figured out is that L = 4 that came from here, we figured out that α ^2 is 4, that came from here there is my α^2.2634

We got BN, we figure that out in example 5, quite a bit of work to figure that out, that is probably the most work in the problem is actually finding those Fourier coefficients and I think that is all we need, we can plug everything in here.2651

I'm going to go ahead and factor out an a/π because that is the same everywhere, a/π from the BN, now we got -1^n + 1/n sin(n π)x/4.2665

Now α ^2 is 4 L^2 is 16 so I can simplify that a little bit, adjust it to 1/4 e^-n^2 π^2, I'm going to combine my 4 and my 16 into the denominator of 4 and that is all I have.2688

That is my complete solution to that boundary value problem that is solution to the differential equation and the 2 boundary conditions and the initial condition and it only took us 4 lectures to get that.2716

Let me recap that, we got α^2 is = 4 here, we essentially done all the hard work of solving this problem in the previous example, if you have not just watched example 5, that is where I got the BN is -1^n +1 x 8/n(π).2737

I did not figure that out right here, that is just not a guess, that came from doing a lot of integration and some work with a Fourier series in the previous problem, having found that.2755

However I can just take that BN and drop that into my generic formula for the solution to the heat equation which we derived several lectures ago, bn sin(n π)x/L e^n^2 π^2 α^2 t^L^2.2764

Our L is 4, we get that from the problem here so we get 16 in the denominator there α ^2 is 4, that is there, 4/16 is is where I got this 4 in the denominator, those terms are both coming from BN and I factored out the 8/π because that did not depend on n.2782

Here I plugged in L=-4 and here I plug in what I know α ^2 is 4 and L ^2 is 16, that is our complete solution to that partial differential equation and to the boundary value problem.2806

That is the end of our lecture on solving the heat equation, you should know how to solve the heat equation now and that actually wraps up this chapter on partial differential equations and that wraps up this whole set of lectures on differential equations.2825

It has been a long journey but I really appreciate your watching these lectures and joining me, I have had a lot of fun talking about differential equations and I hope they go well for you.2842

Just to remind you these are the differential equations lectures on www.educator.com. My name is Will Murray and I really want to thank you for sticking with me through all the lectures, bye.2852

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